Continuous for each variables does not implies continuous

Question

Prove or disprove the following statement:

Statement. Continuous for each variables, when other variables are fixed, implies continuous? More clearly, prove or disprove the following problem:

Let $\displaystyle f:\left[ a,b \right]\times \left[ c,d \right]\to \mathbb{R}$ for which:

• For every $\displaystyle {{x}_{0}}\in \left[ a,b \right]$, $\displaystyle f\left( {{x}_{0}},y \right)$ is continuous on $\displaystyle \left[ c,d \right]$ respect to variable $ \displaystyle y$.
• For every $ \displaystyle {{y}_{0}}\in \left[ c,d \right]$, $ \displaystyle f\left( x,{{y}_{0}} \right)$ is continuous on $ \displaystyle \left[ a,b \right]$ respect to variable $\displaystyle x$.

Then $\displaystyle f\left( x,y \right)$ is continuous on $ \displaystyle \left[ a,b \right]\times \left[ c,d \right]$. ?

https://hongnguyenquanba.wordpress.com/2016/05/12/problem-6/

Answer 1

How about $$f(x,y)=\begin{cases}\frac{xy}{x^2+y^2},&(x,y)\neq(0,0)\\ 0,&(x,y)=(0,0)\\ \end{cases}$$

Answer 2

I have built a counter-example for this statement. Because the building process is quite complicated (it has images too). So I post my counter example in the following link: https://hongnguyenquanba.wordpress.com/2016/05/12/problem-6/

Answer 3

Standard example: $$ f(x,y) = \dfrac{xy}{x^2 + y^2}$$ with $f(0,0) = 0$.